A student was tasked to perform gravimetric analysis of a soluble sulfate. His unknown sample weighed 0.7543 g. The sample underwent precipitation using BaCl2 and was digested for overnight. The precipitate was then filtered off to obtain white crystalline precipitate that was collected in an ash less filter paper. In performing constant weighing, he obtained a crucible mass that is 29.9442 g. After burning his samples inside the crucible, the obtained mass was 30.3375 g.
Questions:
Compute for the experimental mass (g) of SO3 in grams obtained by the student. A) 0.1439 B) 0.1349 C) 0.1943 D) 0.1394 E) 0.3591
Compute for the experimental % SO3 obtained by the student.
A) 73.21 B) 56.33 C) 17.89 D) 56.89 E) 72.80
Compute for the theoretical % SO3 obtained by the student
A) 0.3933 B) 56.37 C) 17.33 D) 17.89 E) 0.4252
Compute for the % error of the student in his analysis.
A) 53.67 B) 68.27 C) 75.43 D) 11.34 E) 42.52
Ah! You just want answers. You don't want to know how to work the problem.
30.3375 g = mass xble + mass ppt
-29.9442 g = mass xble
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....0.3933 = mass ppt = mass SO3
experimental % SO3 in the sample = 0.3933 x (molar mass SO3/molar mass BaSO4)*100 = ? %SO3
I don't know how to answer the last two because I don't know the %SO3 in the unknown. You are looking for theoretical % and % error and I don't know how you calculate that without knowing % SO3 in the unknown. I answered a similar question a few days ago and I had the same problem. The trouble may be because of the word "theoretical" so it may be a language problem.
To compute for the experimental mass of SO3 obtained by the student, we need to calculate the difference in mass before and after burning the sample. Here's how you can do it:
1. Calculate the mass of the precipitate obtained by subtracting the mass of the empty crucible from the mass of the crucible with the precipitate:
Mass of precipitate = Crucible mass after burning - Crucible mass before burning
Mass of precipitate = 30.3375 g - 29.9442 g
2. Now, you need to convert the mass of the precipitate to the mass of SO3. Since the precipitate is composed of BaSO4 (barium sulfate), the molar mass of SO3 is 80.07 g/mol. Convert the mass of the precipitate to moles of SO3:
Moles of SO3 = Mass of precipitate / Molar mass of SO3
Moles of SO3 = (30.3375 g - 29.9442 g) / 80.07 g/mol
3. Finally, convert the moles of SO3 to grams by multiplying it by the molar mass of SO3:
Experimental mass of SO3 = Moles of SO3 * Molar mass of SO3
Now you can calculate the experimental mass of SO3 using the given values.
To compute for the experimental % SO3 obtained by the student, divide the experimental mass of SO3 obtained by the student by the initial mass of the sample and multiply by 100.
To compute for the theoretical % SO3 obtained by the student, you need the stoichiometry of the reaction between the original sulfate compound and BaCl2 to form the precipitate. Once you have the stoichiometry, you can calculate the theoretical mass of SO3 using the molar mass.
To compute for the % error of the student in his analysis, use the formula:
% error = |(experimental % SO3 - theoretical % SO3) / theoretical % SO3| * 100
Now you can compute for the answers to the given questions using the steps and formulas provided.